Natural Number Multiplication Distributes over Addition

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Theorem

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:

$\forall x, y, z \in \N:$
$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$


Proof 1

\(\ds \paren {x + y} \times z\) \(=\) \(\ds +^z \paren {x + y}\) Definition of Natural Number Multiplication
\(\ds \) \(=\) \(\ds \paren {+^z x} + \paren {+^z y}\) Power of Product of Commuting Elements in Semigroup equals Product of Powers
\(\ds \) \(=\) \(\ds x \times z + y \times z\)

$\Box$

\(\ds z \times \paren {x + y}\) \(=\) \(\ds +^{x + y} z\) Definition of Natural Number Multiplication
\(\ds \) \(=\) \(\ds \paren {+^x z} + \paren {+^y z}\) Index Laws for Semigroup: Sum of Indices
\(\ds \) \(=\) \(\ds z \times x + z \times y\)

$\blacksquare$


Proof 2

We are to show that:

$\forall x, y, z \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$


From the definition of natural number multiplication, we have by definition that:

\(\ds \forall m, n \in \N: \, \) \(\ds m \times 0\) \(=\) \(\ds 0\)
\(\ds m \times n^+\) \(=\) \(\ds \paren {m \times n} + m\)


Let $x, y \in \N$ be arbitrary.

For all $z \in \N$, let $\map P z$ be the proposition:

$\forall x, y \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \paren {x + y} \times 0\) \(=\) \(\ds 0\) Definition of Natural Number Multiplication‎
\(\ds \) \(=\) \(\ds 0 + 0\) Definition of Natural Number Addition
\(\ds \) \(=\) \(\ds x \times 0 + y \times 0\) Definition of Natural Number Multiplication‎

and so $\map P 0$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.


So this is our induction hypothesis:

$\forall x, y \in \N: \paren {x + y} \times k = \paren {x \times k} + \paren {y \times k}$


Then we need to show:

$\forall x, y \in \N: \paren {x + y} \times k^+ = \paren {x \times k^+} + \paren {y \times k^+}$


Induction Step

This is our induction step:


\(\ds \paren {x + y} \times k^+\) \(=\) \(\ds \paren {x + y} \times k + \paren {x + y}\) Definition of Natural Number Multiplication‎
\(\ds \) \(=\) \(\ds \paren {x \times k} + \paren {y \times k} + \paren {x + y}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {\paren {x \times k} + x} + \paren {\paren {y \times k} + y}\) Natural Number Addition is Commutative and Associative
\(\ds \) \(=\) \(\ds \paren {x \times k^+} + \paren {y \times k^+}\) Definition of Natural Number Multiplication

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:

$\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$

$\Box$


Next we need to show that:

$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

So:

\(\ds z \times \paren {x + y}\) \(=\) \(\ds \paren {x + y} \times z\) Natural Number Multiplication is Commutative
\(\ds \) \(=\) \(\ds \paren {x \times z} + \paren {y \times z}\) from above
\(\ds \) \(=\) \(\ds \paren {z \times x} + \paren {z \times y}\) Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

$\blacksquare$


Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

$\forall x, y, z \in \N_{> 0}:$
$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$


Using the following axioms:

\((\text A)\)   $:$     \(\ds \exists_1 1 \in \N_{> 0}:\) \(\ds a \times 1 = a = 1 \times a \)             
\((\text B)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds a \times \paren {b + 1} = \paren {a \times b} + a \)             
\((\text C)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds a + \paren {b + 1} = \paren {a + b} + 1 \)             
\((\text D)\)   $:$     \(\ds \forall a \in \N_{> 0}, a \ne 1:\) \(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \)             
\((\text E)\)   $:$     \(\ds \forall a, b \in \N_{> 0}:\) \(\ds \)Exactly one of these three holds:\( \)             
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)             
\((\text F)\)   $:$     \(\ds \forall A \subseteq \N_{> 0}:\) \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)             


Left Distributive Law for Natural Numbers

First we show that:

$n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$


Let us cast the proposition in the form:

$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$


Basis for the Induction

$\map P 1$ is the case:

\(\ds a \times \paren {b + 1}\) \(=\) \(\ds \paren {a \times b} + a\) Axiom $\text B$
\(\ds \) \(=\) \(\ds \paren {a \times b} + \paren {a \times 1}\) Axiom $\text A$

and so $\map P 1$ holds.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$


Then we need to show:

$\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$


Induction Step

This is our induction step:

\(\ds a \times \paren {b + \paren {k + 1} }\) \(=\) \(\ds a \times \paren {\paren {b + k} + 1}\) Axiom $\text C$
\(\ds \) \(=\) \(\ds \paren {a \times \paren {b + k} } + a\) Axiom $\text B$
\(\ds \) \(=\) \(\ds \paren {\paren {a \times b} + \paren {a \times k} } + a\) Induction hypothesis
\(\ds \) \(=\) \(\ds \paren {a \times b} + \paren {\paren {a \times k} + a}\) Natural Number Addition is Associative
\(\ds \) \(=\) \(\ds \paren {a \times b} + \paren {a \times \paren {k + 1} }\) Axiom $\text B$

The result follows by the Principle of Mathematical Induction.

$\Box$


Right Distributive Law for Natural Numbers

Then we show that:

$\paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$


For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \paren {a + b} \times 1\) \(=\) \(\ds a + b\) Axiom $\text A$
\(\ds \) \(=\) \(\ds \paren {a \times 1} + \paren {b \times 1}\) Axiom $\text A$

and so $\map P 1$ holds.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$


Then we need to show:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$


Induction Step

This is our induction step:

\(\ds \paren {a + b} \times \paren {k + 1}\) \(=\) \(\ds \paren {\paren {a + b} \times k} + \paren {a + b}\) Axiom $\text B$
\(\ds \) \(=\) \(\ds \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}\) Induction hypothesis
\(\ds \) \(=\) \(\ds \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}\) Natural Number Addition is Commutative
\(\ds \) \(=\) \(\ds \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }\) Axiom $\text B$

The result follows by the Principle of Mathematical Induction.

$\Box$

The result follows.

$\blacksquare$


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